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Correctly calculating gravity classically. 
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#1
Sep611, 05:40 PM

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I'm someone who's always been a person who has casually interested in math. Recently, after a long break, I've regained a bit of this interest and looked into how gravity is calculated. I'm sure most know the G*m1*m2/r^2, and while a good approximation of gravity, it is not completely correct. We see this when the real world has slight differences from what is expected by this formula and what actually happens. I gave a great deal of thought to why this formula finally falls short and came to an answer. We make an assumption by assuming that calculating by the center of mass has no effect on how gravity interacts. I tested this theory by comparing the results of multiple gravitating forces on each other, and how it differs from if just one object had been gravitating. There is no part in this formula which accounts for the size and mass distribution of objects.
So we come to part two, I thought about how gravity would need to be calculated in order to account for this difference. The answer is a rather interesting and simple idea. At any equal distance from a central point, all mass will have an equal effect of gravity, simply in different vectors. Why not instead of thinking of gravity as a force between two points, think of it as the sum of the surface of all spheres 0 to infinity from your central point. Now, we just need to figure out, what is the gravitational that each sphere's surface will have? Surpringly simple, if the Center of mass for this surface is R(x) distance from the center, and surface is at a radius of x, and the mass of this surface is M(x), the formula is: F= G*M1*M(x)*R(x)/x^3 applied at vector of M1>R(x) So the total gravitational force on any object for any number of other objects at any distances, would be the sum 0 to infinity of said object. Substitute in points, and this gives the same answer as newtonian physics as R(x)/x^3 will cancel to 1/x^2 if R(x) and x are equal. And finally, part 3. Albeit exceedingly more complex, since you need to develop a formula to describe the distribution of mass in the system, why isnt a formula like this used for classical gravitation? It predicts small changes from standard classical models with a quick overview. The biggest one is that objects that get very close have their gravitational pull in a way 'dispersed' into less coherent directions. This would especially be noticeable in objects with highly eccentric orbits. Gravitation in these situations would be slightly less than expected in newtonian, and approach but never quite reach newtonian as objects get far. This explains mercury's precession, and explains the pioneer anomaly in classical mechanics. I'd love to hear comments from someone more knowledgeable, as I have an entirely self taught knowledge on this subject. Please let me know if I need to give examples. Thanks for reading. 


#2
Sep611, 05:49 PM

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[quote] We see this when the real world has slight differences from what is expected by this formula and what actually happens. I gave a great deal of thought to why this formula finally falls short and came to an answer. We make an assumption by assuming that calculating by the center of mass has no effect on how gravity interacts.[\quote] What? On the contrary, the "error" in that approximation is in assuming that the center of mass is the ponly thing that counts! 


#3
Sep611, 05:57 PM

P: 21

ST  Frankly, I do not know what you are talking about. Newtonian theory simply accommodates the behavior of one accumulated mass in the presence of another accumulated mass. It is accurate enough to explain all visible planetary motion except that of Mercury. Mercury is close enough to the sun that relativistic calculations are required.



#4
Sep611, 06:20 PM

P: 2

Correctly calculating gravity classically.



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